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Proof step 16 from proof wand-shift (previous step)

P0(H1), P1(H), P2(H ∪ H0), H ⊥ H0, H ⊥ H1, H1 ⊥ H0 ⊦ H1 ⊥ H ∪ H0 ∧ H ∪ H1 ∪ H0 = H1 ∪ H ∪ H0 ∧ P0(H1) ∧ P2(H ∪ H0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
P0(H1), P1(H), P2(H ∪ H0), H1 ⊥ H0, H ⊥ H1, H ⊥ H0 ⊦ H ∪ H1 ∪ H0 = H1 ∪ H ∪ H0

Rule argument

Additional information

Used simplifier rules: ⊦ H0 ⊥ H1 → H1 ⊥ H0; (s)% ⊦ H1 ⊥ H → (H0 ⊥ H1 ∪ H ↔ H0 ⊥ H1 ∧ H0 ⊥ H); (ws)